Chemistry

Propanoic acid has a Ka of 1.3 x 10^-5. What is the % ionization in a 3.0M solution?

asked by Lily
  1. Let's call propanoic acid HPr. Then
    ...........HPr ==> H^+ + Pr^-
    Initial....3.0......0.....0
    Change......-x......x......x
    Equil.....3.0-x.....x.......x

    Ka = (H^+)(Pr^-)/(HPr)
    Substitute from the ICE chart into the Ka expression and solve for x = (H^+).
    Then %ionization = [(H^+)/3.0)]*100 = ?

    posted by DrBob222
  2. thank you so much!

    posted by Lily

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